The Ka of a monoprotic weak acid is 5.81 × 10-3. What is the percent ionization of a 0.111 M solution of this acid?My calculations:Ka= 5.81x10^-3=x^2/0.111---> x^2= (5.81x10^-3)(0.111)---->x^2=6.4991x10^-4----> x= 0.0254Percent Ionization= (0.0254/0.111)x100%= 22.88%As you can see I was only able to solve this halfway through. It's not valid because it's more than 5% so I have to use the quadratic formula. Problem is I don't know what values to place as a,b, or c for the quadratic formula. Can anyone please help me with this. The best explanation gets 10 points!. Created by MJ. science-mathematics-en - chemistry-en

The Ka of a monoprotic weak acid is 5.81 × 10-3. What is the percent ionization of a 0.111 M solution of this?

MJ @MJ

May 2021 2 101 Report

The Ka of a monoprotic weak acid is 5.81 × 10-3. What is the percent ionization of a 0.111 M solution of this?

The Ka of a monoprotic weak acid is 5.81 × 10-3. What is the percent ionization of a 0.111 M solution of this acid?

My calculations:

Ka= 5.81x10^-3=x^2/0.111---> x^2= (5.81x10^-3)(0.111)---->x^2=6.4991x10^-4----> x= 0.0254

Percent Ionization= (0.0254/0.111)x100%= 22.88%

As you can see I was only able to solve this halfway through. It's not valid because it's more than 5% so I have to use the quadratic formula. Problem is I don't know what values to place as a,b, or c for the quadratic formula. Can anyone please help me with this. The best explanation gets 10 points!

Science & Mathematics / Chemistry

Atheists Are Superior

Verified answer

5.81e-3 = x*x / 0.111-x

6.449e-4 - 5.81e-3x - x^2 = 0

A = -1

B = -5.81e-3

C = +6.449e-4

Plug it in the quadratic equation formula

x = 0.022655

% Ionization = (0.022655/0.111) * 100 = 20.41%

So not much difference. Your method is correct

Read up on some info about Quadratic Formula

http://en.wikipedia.org/wiki/Quadratic_equation

childress

anticipate a susceptible monoprotic acid HA. HA ? H(+) + A(-) at t=0 C 0 0 at eq. C(a million-x) Cx Cx Now, [HA] initially is C(=0.174 M). If 'x' is the degree of dissociation of acid, at equilibrium, [HA] remaining will be C-Cx, i.e. C(a million-x), and [H(+)] & [A(-)] shaped should be Cx. Now, Ka = [A(-)][H(+)]/[HA] => Ka = Cx²/(a million-x) Now, on condition that Ka = 2.5×10^-3 and for on account that answer is dilute, we may be able to approximate (a million-x) as a million. therefore, inducing the above reported consequences and value of C as 0.174M, we get 2.5×10^-3 = 0.174x² => x ? 0.11987 therefore, % ionisation = 11.987% ? 12% it is also possible to analyze the end result by utilizing no longer utilizing the approximation. if so a quadratic equation will be shaped. you receives 2 roots, one helpful and one unfavourable. forget the unfavourable one. by utilizing that technique, you get x ? 0.1129